AI HL · 12.1
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Mixed Practice & Exam Prep
12.1.6
€4200 is invested in an account that pays 3.5% simple interest each year. Find the amount in the account after 6 years.
Have a go first. Then open the working one line at a time.
\(3.5\%\) of \(4200 = 147\)
Simple interest adds the same amount every year, so the amounts form an arithmetic sequence.
\(4200 + 6 \times 147 = 5082\)
Six lots of €147 are added.
Plan A pays $30 000 in the first year and $1500 more each year. Plan B pays $30 000 in the first year and 4% more each year. In which year does Plan B first pay more than Plan A?
Have a go first. Then open the working one line at a time.
\(A_n = 30\,000 + 1500(n - 1)\)
Adding the same amount: arithmetic.
\(B_n = 30\,000 \times 1.04^{n-1}\)
Multiplying by the same number: geometric.
Year 12: A 46 500, B 46 183.62. Year 13: A 48 000, B 48 030.97.
Plan B first pays more in year 13.
A museum records its visitors on five days: 1240, 1310, 1395, 1450, 1520. Modelling the data as an arithmetic sequence, predict the number of visitors on day 9.
Have a go first. Then open the working one line at a time.
\(d = \frac{1520 - 1240}{4} = 70\)
The differences are not all the same, so use the first and last values. Five values have four gaps.
\(u_9 = 1240 + (9 - 1)(70) = 1800\)
Eight steps of 70 from day 1.
A gym models its number of members in month \(n\) by \(u_n = 240 + 15(n - 1)\). In month 10 the gym actually has 392 members. Find the percentage error in the model's prediction, to 3 significant figures.
Have a go first. Then open the working one line at a time.
\(u_{10} = 240 + 15(10 - 1) = 375\)
Put \(n = 10\) into the model.
\(\left|\frac{375 - 392}{392}\right| \times 100 = 4.337\ldots\)
The difference, over the actual value, times 100.
\(4.34\%\)
The model predicts 17 members too few.
A lake has 1500 fish in year 1. A model says the number of fish grows by 12% each year. (a) Find the number of fish the model predicts for year 40, to 3 significant figures. (b) Comment on whether the model is sensible for year 40.
Have a go first. Then open the working one line at a time.
\(u_n = 1500 \times 1.12^{n-1}\)
Multiplying by 1.12 each year: geometric.
\(u_{40} = 1500 \times 1.12^{39} = 124\,621.8\ldots\)
About 125 000 fish.
Not sensible. The model keeps multiplying by 1.12, so by year 40 it predicts about 83 times as many fish as in year 1. A real lake has limited food and space, so the number of fish would level off.
Say what the model does (keeps multiplying), say what happens in the real situation, and link the two. "It is too big" on its own is not enough.
A donor offers a charity two plans, each lasting 10 years. Plan A gives $400 in year 1 and $60 more each year. Plan B gives $400 in year 1 and 9% more each year. Which plan gives the charity more in total? Justify your answer.
Have a go first. Then open the working one line at a time.
\(S_{10}\) for each plan
"In total" means a sum, not the gift in year 10.
\(S_{10} = \frac{10}{2}[2(400) + (10 - 1)(60)] = 6700\)
Adding $60 each year: arithmetic.
\(S_{10} = \frac{400(1.09^{10} - 1)}{1.09 - 1} = 6077.17\)
Multiplying by 1.09 each year: geometric.
$6700 against $6077.17
Plan A gives more in total, by $622.83. A justification needs both values and a comparison, not just the name of a plan.
Stage 1 of a pattern has 1 shaded triangle, stage 2 has 3 and stage 3 has 9. Each shaded triangle is replaced by 3 smaller ones at the next stage. (a) How many shaded triangles are in stage 8? (b) Find the total number drawn in stages 1 to 8. (c) In which stage does the total number drawn first pass 10 000?
Have a go first. Then open the working one line at a time.
\(1,\ 3,\ 9,\ \ldots\)
Multiplying by 3 each time: geometric with \(u_1 = 1\) and \(r = 3\).
\(u_8 = 1 \times 3^{7} = 2187\)
Seven multiplications by 3.
\(S_8 = \dfrac{3^8 - 1}{3 - 1} = 3280\)
The total drawn in stages 1 to 8.
\(S_9 = 9841\), \(S_{10} = 29524\)
Use a table. Stage 10 is the first with a total above 10 000.